Infinite square root

Date: 6/4/96 at 13:29:14
From: Anonymous
Subject: Infinite square root
I know this is an easy question, but I do not know how to figure it
out.
For instance, if y = sqrt(2+ sqrt(2+ sqrt(2+ sqrt(2+ ..., y = 2, and
this happens always, because if k = e+ sqrt(e+ sqrt(e+ ..., and it
goes indefinitely, k = e. However, how can I prove that this is true,
using normal properties of roots?

Date: 6/4/96 at 16:14:46
From: Doctor Darrin
Subject: Re: Infinite square root
If we think about the properties that y has to have, we can find its
value.
What do we get if we square y? We get 2+sqrt(2+sqrt(2+..., so we see
that y^2-2=sqrt(2+sqrt(2+sqrt(2+... . Thus, y^2-2=y, so y is a root
of the quadratic equation y^2-y-2==0. You can work out what the roots
of this equation are, and this gives two possible values of y (one is
positive and one is negative). Since y is defined as a square root,
it makes sense to take the positive value.
I am not sure what you mean when you say that "this happens always".
In the definition you give for y, if we replace 2 by another number
(say n), we can use essentially the same procedure to find what y is,
but in general, y does not equal n. When n=3, y=2.302..., (the
positive root of y^2-y-3=0) and when n=4, y=2.56... The definition
you give for k is different, and in fact k does not equal e in
general. For e=3, k=5.302... In fact, there is no value of e (except
e=0) for which k=e.
Thanks for asking the question, it is very interesting
-Doctor Darrin, The Math Forum
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